Online calculator to calculate the sum of the terms in a geometric sequence. If A1, A2, ... , An, ... is a geometric sequence with common ratio r, this calculator calculates the sum Sn given by Sn = A1 + A2 + ... + An = a1(1-rn)/(1-r)and the nth term an = a1 rn - 1 Use of the Geometric Series calculator1 - Enter the first term A1 in the sequence, the common ratio r and n n the number of terms in the sum then press enter. A1 and r may be entered as an integer, a decimal or a fraction. n must be a positive integer. More References and LinksMath Calculators and Solvers. |
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Calculate series and sums step by step
This calculator will try to find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). It will also check whether the series converges.
Answer
Your input: calculate $$$\sum_{n=1}^{\infty} 3^{- n}$$$
$$$\sum_{n=1}^{\infty} 3^{- n}$$$ is an infinite geometric series with the first term $$$b=\frac{1}{3}$$$ and the common ratio $$$q=\frac{1}{3}$$$.
By the ratio test, it is convergent.
Its sum is $$$S=\frac{b}{1-q}=\frac{1}{2}$$$.
Therefore,
$$\color{red}{\left(\sum_{n=1}^{\infty} 3^{- n}\right)}=\color{red}{\left(\frac{1}{2}\right)}$$
Hence,
$$\sum_{n=1}^{\infty} 3^{- n}=\frac{1}{2}$$
Answer: $$$\sum_{n=1}^{\infty} 3^{- n}=\frac{1}{2}$$$
Infinite Geometric Series Formula: s∞=a1−r
Enter the value of First term(a)=
Enter the value of Common ratio(r)=
s∞=
Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. BYJU’S online infinite geometric series calculator tool makes the calculation faster, and it displays the sum in a fraction of seconds.
How to Use the Infinite Geometric Series Calculator?
The procedure to use the infinite geometric series calculator is as follows:
Step 1: Enter the first term and common ratio in the respective input field
Step 2: Now click the button “Calculate” to get the sum
Step 3: Finally, the sum of the infinite geometric sequence will be displayed in the output field
What is Meant by Infinite Geometric Series?
In Mathematics, the infinite geometric series gives the sum of the infinite geometric sequence. It has the first term (a1) and the common ratio(r). It has no last term. If the common ratio of the infinite geometric series is more than 1, the number of terms in the sequence will get increased. In this case, if you try to add larger numbers many times, the series will result in infinity. The general term to represent the infinite geometric sequence is given by as a+ar+ar2+ ar3+ ….
Hence, the sum of the infinite geometric series with the common ratio -1<r< 1 is given by:
S = a1/(1-r)
A geometric series is a sequence of the form
ra⁰, ra¹, ra², ra³,...
for some numbers r and a. In a geometric series, the ratio of consecutive terms is constant. Such sequences appear in discrete math problems such as compound interest, or processes with constant growth/decay rates. One of nice properties of geometric functions is that they are easy to sum. And if a is between -1 and 1, the infinite sum converges.
The summation formula is explained below, or you can use the calculator on the left. To use the calculator, enter the value of a as a fraction and the summation index values. For the infinite sum, enter "infinity" in the field for the upper index.
Formula for the Sum of a Geometric Series
To find the sum of a finite number of terms in a geometric series, consider the expression
G = a⁰ + a¹ + a² + ... + an
If we multiply both sides by a, we have
aG = a¹ + a² + a³ + ... + an + an+1
= (a⁰ + a¹ + a² + ... + an) - a⁰ + an+1
= G - 1 + an+1
Solving this expression for G gives us
G = (an+1 - 1)/(a - 1).
If |a| is less than 1 and n goes to infinity, the infinite sum is -1/(a-1), or 1/(1-a).
Example 1:
Compute the sum
3 + 3(0.25) + 3(0.25)² + ... + 3(0.25)²⁰
First factor out the 3, since it can be multiplied at the end. This leaves us with
1 + 0.25 + 0.25² + ... + 0.25²⁰
= (0.25²¹ - 1)/(0.25 - 1)
= (1 - 0.25²¹)/0.75
= (4/3)[1 - (1/4)²¹]
Returning the factor of 3 gives us the final answer of
4[1 - (1/4)²¹]
Example 2:
Let c = 1/π ≈ 0.31831. Simplify the infinite sum
B = c - c³ + c⁵ - c⁷ + c⁹ - ...
First, divide both sides by c:
B/c = 1 - c² + c⁴ - c⁶ + c⁸ - ...
This is equivalent to
B/c = (-c²)⁰ + (-c²)¹ + (-c²)² + (-c²)³ + (-c²)⁴ + ...
So we have
B/c = 1/[1 - (-c²)]
B/c = 1/[1 + c²]
B = c/[1 + c²]
B = (1/π)/[1 + 1/π²]
B = π/(1 + π²)
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